Elementary examples of pointwise sequential closures that are not pointwise closed

Question

For a topological space $X$ let $\mathcal{F}$ be a set of functions taking $X$ to $\mathbb{R}$. Are there elementary examples of $\mathcal{F}$ whose pointwise sequential closure $\operatorname{scl}(\mathcal F)$ is not pointwise closed?

Or equivalently, $$ \operatorname{scl}(\mathcal F) \subsetneqq \operatorname{scl}(\operatorname{scl}(\mathcal F)).\qquad \qquad(*) $$

I may be pushing the idea of 'elementary' — I'm thinking more or less the level of freshman calculus. The theory of Baire classes enumerates a hierarchy of classes of $\mathbb{R} \to \mathbb{R}$ functions, where class $n+1$ is the pointwise closure of class $n$, and class $0$ consists of the continous functions. So each Baire class is an example of (*). In my opinion, Baire classes go beyond elementary, but posters can use whatever definition of elementary they are comfortable with.

Answer 1

By definition the pointwise closure is pointwise closed. But your example of Baire functions suggests that you're actually thinking of sequential pointwise closure, which gives a (possibly) smaller set than the actual pointwise closure. This is as $\mathbb{R}^X$ is not a sequential space, for many uncountable $X$, certainly for $X$ of size $\mathfrak{c}$, possibly earlier.

Answer 2

In the following, $\mathbb{N}^+$ denotes the positive integers , and $\mathbb{N}^{\mathbb{N}^+}$ is the set of all sequences of natural numbers. All limits and closures are pointwise.

Let the set $X$ be uncountable.

Let $\varphi:\mathbb{N}^{\mathbb{N}^+} \to X$ be one-to-one.

Let $\mathcal{F}=\{f_{i,j}: X \to \mathbb{R} \mid i,j \in \mathbb{N}^+\}$ where

$$ f_{i,j}(x) = \begin{cases} 1 & \text{if $g=\varphi^{-1}(x)$ exists and $g(i)=j$,} \\ 1/i & \text{otherwise.} \end{cases} $$

It is easy to verify that $\lim_{j \to \infty} f_{i,j} = 1/i$ and $\lim_{i \to \infty} 1/i = 0$. (Here, $1/i$ and $0$ denote constant functions.)

If we can show that the constant function $0$ is not a pointwise limit of $\mathcal{F}$, we are done. Indeed, suppose that there is some sequence $\{i_k,j_k\}$ such that $\lim_{k \to \infty} f_{i_k,j_k} = 0$. We can assume that $\{i_k\}$ is strictly increasing. Then define $\rho: \mathbb{N}^+ \to \mathbb{N}$ as follows: $$ \rho(n) = \begin{cases} j_k & \text{if $n=i_k$ for some $k$,} \\ 0 & \text{otherwise.} \end{cases} $$

Then $\lim_{k \to \infty} f_{i_k,j_k}(\varphi(\rho)) = 1$, which is a contradiction.

As a visual aid, we can write out the elements of $\mathcal{F}$ as $$ \begin{matrix} f_{1,1} & f_{1,2} & f_{1,3} & \cdots & f_{1,j} & \cdots \\ f_{2,1} & f_{2,2} & f_{2,3} & \cdots & f_{2,j} & \cdots \\[1ex] \vdots & \vdots& \vdots & \ddots & \vdots \\ f_{i,1} & f_{i,2} & f_{i,3} & \cdots & f_{i,j} &\cdots \\[1ex] \vdots & \vdots& \vdots & \cdots & \vdots \\ \end{matrix}. $$

If we follow across along a row $i$ the $f_{i,j}$ will converge to $1/i$. However, if we try to go downwards, no matter which path we take on the way to the limit $0$, we will always be frustrated at some $x \in X$ where we converge to $1$ instead of $0$. (So in this example $\varphi$ stands for "$\varphi$rustration".)

As a final remark, note that $X$ must be uncountable for this construction to work (because $\mathbb{N}^{\mathbb{N}^+}$ is uncountable). So this example will not work for functions $\mathbb{Q} \to \mathbb{R}$.

Answer 3

Let $I=[0,1]$ and let $\mathcal{F}$ be the continuous functions $I \to \mathbb{R}$. Let $1_\mathbb{Q}$ be the function on $I$ that is $1$ on rationals and $0$ otherwise. We claim that $1_\mathbb{Q} \in \operatorname{scl}(\operatorname{scl}(\mathcal F)) \setminus \operatorname{scl}(\mathcal F)$.

Elsewhere on this site, answer https://math.stackexchange.com/a/1444407/1257 gives an elementary proof that there is no sequence of continuous functions that converges to $1_\mathbb{Q}$. But if $(q_i)$ is an enumeration of the rationals in $I$, we can build sequences of continous functions that pointwise converge to $\chi_{\{q_i|i \le n\}}$, and these limit functions in turn pointwise converge to $1_\mathbb{Q}$.